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Can the use of wide-chord supercritical airfoil blades in a helicopter with coaxial counter-rotating rotors solve retreating blade stall?
The problem with retreating blade stall is that it limits the VNE and the overall top speed of helicopters. Retreating blade stall is basically the tip of the retreating blade having too much AoA at a lower speed, causing it to stall.
In general, coaxial counter-rotating helos are faster and mitigate retreating blade stall effects (they experience increased drag). Supercritical airfoils are designed to operate at far higher speeds than conventional airfoils (thus can produce more lift), have a high maximum lift and docile stall behavior, and generally stall at higher AoA.
Would this combination allow to reduce or eliminate retreating blade stall?
There have also been a couple of studies on the use of supercritical airfoils on helos. Here's one:
https://www.techscience.com/iasc/v33n1/46153/html

Counter-rotating rotating rotors is enough to effectively eliminate the retreating blade stall problem.  Note that the CH-47 is considered the fastest helicopter in the US military.
Retreating blade stall is strongly related to a requirement for roll equilibrium.  The left half of the rotor needs to create as much lift as the right half -- otherwise, the aircraft will roll.
The retreating blade is less able to produce this lift (as it has lower effective velocities) and so (on a single-rotor helicopter) you must increase the angle of attack on the retreating blade to make this possible.  This blade therefore stalls before the advancing blade.  This dramatically increases the power required to turn the rotor and limits the speed the aircraft can achieve.
Any helicopter with left-right symmetrical lifting systems (i.e. counter rotating rotors) can balance the lift from one rotor's advancing blade with the other rotor's advancing blade.  This allows you to set the angle of attack of the retreating blade to the minimum drag setting.  The retreating blades will not contribute to lift, but they also will not cause an early dramatic rise in drag.

Wasn't the Westland Lynx the one holding the world record for straight flying?

So a supercritical airfoil could be detrimental or helpful?

Also, what are possible solutions to Dynamic and Static rollover, Ground resonance, Dynamic stall and Vortex ring state? (from a design standpoint)

What about adding winglets or using blade shapes like airbus' blue edge blades? Or something similar like the scimitar shaped fan blades in the GE90?

@BanzaiFighterbomber: you'd better post these comments as questions

I think the Lynx with the record was substantially modified from the version in service with the British military.  The CH-47 comparisons I've seen probably restrict it to comparisons within the US military.  Both the Lynx (201) and CH-47 (200) appear to have essentially equivalent to speeds, with one perhaps pushing a little more for bragging rights.

Supercritical airfoils are unsuitable for helicopter blades due to their high pitch moment coefficients.
The idea of a supercritical airfoil is to separate the effects of thickness and lift creation on its pressure distribution. There is more to them, so this is probably an oversimplification, but it should help to understand why supercriticality is connected to high pitching moments.
Those pitching moments will twist the high aspect ratio helicopter blades in uncontrollable ways. Given the variation in lift over the rotor disk, the consequent variation in the local center of pressure will make it impossible to control the local angle of attack of the blade. Only with pitch-neutral airfoils like the venerable NACA 8H12 or the Hughes HH02 will a helicopter rotor become controllable.
Yes, there are papers out there which claim an advantage for supercritical airfoils. But when I read sentences in them like this one ("The supercritical airfoils are specially used in the supersonic aircraft to cross the sonic barrier"), the authors are evidently clueless and I wonder how this nonsense survived a peer review.

Can this be solved by using washout?

@BanzaiFighterbomber: washout is abundantly used on helicopter blades

"and I wonder how this nonsense survived a peer review" they even forgot to properly attribute the figure they used (Helicopters aerodynamic by Leishman) so I don't even know if this went through a peer review actually...

Supercritical airfoils are designed to operate at far higher speeds than conventional airfoils (thus can produce more lift), have a high maximum lift and docile stall behavior, and generally stall at higher AoA.

These premises about supercritical airfoils in respect to conventional airfoils are not 100% correct:

They are not designed to operate at far higher speed than conventional airfoil rather to retard the generation of shock waves and therefore the relevant drag increase; they generate less drag at higher Mach number but the flying speed is only determined by the available power; most of the jet fighters out there (F16, F18, F22,...) use quite standard NACA airfoils but they can happily fly supersonic.
They do not have a higher maximum $C_l$; being thin and with a small camber, the opposite is actually true, they have a lower maximum $C_l$.
And for the same reason they also stall at lower AoA.
Stall is indeed docile but at the expense of a higher pitching moment which might limit the manoeuvrability and gives for sure high control loads.

Supercritical airfoils (with a bit of tweak of the camber) can and are actually used for the tip of the blades (last 10 to 15% of the bladespan) but not to mitigate the retracting blade stall phenomenon rather to reduce drag at the high Mach numbers typical of the advancing blade. The rest of the blade is designed around a more thick and cambered airfoil.
Please note that the paper you linked tested the airfoils at Mach number "0.3, 0.4 and 0.5" i.e.quite far from the actual conditions seen by the tip of the blade, where the Mach number can be as low as 0.1 on the retracting blade and as high as 0.8 on the advancing blade... They also forgot to calculate drag and pitching moment which are obviously also of paramount importance in blades design: they claim for example that "the rotor thrust is increased about 5% to 10%"; this is a very good result... but what if drag increases of 15%?

Depends on the profile though, NASA SC(2)-0714 doesn't have a thin camber at all.

@BanzaiFighterbomber The NASA SC(2)-0714 is not a helicopter airfoil.

Doesn't mean it can't be used on one. It's the most extensively tested supercritical airfoil. Either on a plane or a helo, still answers the same aerodynamics

@BanzaiFighterbomber: nope, the aerodynamic environment seen by a rotating wing is completely different than a fixed wing. Just the fact that a supercritical airfoil stalls before a conventional airfoil and generates a higher pitching moment is a no-go for its use on a blade (except maybe on its tip but after having tuned its camber)

@sophit        Doesn't make any sense. None of the meaningfully tested supercritical airfoils is thin. None of the Sikorsky airfoils has a small camber, like the SC2110 and SC1095.

@BanzaiFighterbomber: never said that Sikorsky airfoils have small camber, actually the opposite should be true for the reason I explained. Don't know where you've got that...


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Can vortex ring state (VRS) be solved by shaping the blades so that the vortexes they produce are weak enough not to form a vortex ring?
Vortex Ring State is one of the hazardous states a helicopter can find itself into. VRS happens when a helicopter has low airspeed (and usually hovering OGE) and is descending. If the helicpoter is descending too fast (that might be around 300 fpm for some helos) the rotor will enter its own downwash. As this downwash is turbulent, the rotor will be unable to provide enough lift and the helo will start falling.
Would blade shaping (maybe like Airbus' Blue Edge or something similar to the scimitar shaped fan blades of the GE90) help with VRS? Or if not solve it, make it weak enough not to pose a threat.

No.  VRS isn't about the tip vortices you are thinking of.
VRS is a thing even if you had a pure actuator disk.  I.e. it is a fundamental phenomena of powered lift.  It has everything to do with disk loading and just about nothing to do with details of the blade design.

Vortex ring state (VRS) is a blade root stall phenomenon that occurs in rapid descents. In a normal rapid descent, there is a small amount of airflow UP through the center of the rotor disc right at the blade roots.
When the conditions are right, the root stalled section expands out from the hub, and you get a very large upflow near the center of the rotor disc in addition to the up flow beyond the edges of the rotor disc.
This sets up an inner vortex that is more or less a mirror image to the vortex flowing around the blade tips. The two vortex fields meet over the mid span part of the blade and flow down through the disc.
All this downward flow, originating from both the hub area AND the blade tips is too much, and increasing sink rate just makes it worse. You could call it the "Double Donut Of Death" with an outer vortex and an inner vortex both combining their flows to hurtle you toward the ground.
The only way out is to disrupt the vortex somehow, usually by getting some lateral motion going, translating out of the vortex.
Wikipedia covers it pretty well, and once you see what is happening, it becomes apparent that trying to reduce the strength of the vortices at the blade tips with something like a droop tip or winglet (structurally problematic) has little effect, partly because the vortex field extends way beyond the tip, and mostly because the problem originates at the other end of the blade, near the hub.

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What are the effects of whiteout (snow) and brownout (sand/dust) on helos?
Is there any other effect besides low visibility?
Both seem to show very similar conditions.

Effects from the pilot's perspective are roughly the same. The difference between snow and sand is mostly due to the abrasive effects of sand/dust on rotor blades and especially engines.
A helicopter operating in the desert will have a much shorter engine life from sand erosion of compressor/turbine blades, and shorter main and tail rotor blade life, compared to one operating in the arctic that is physically unaffected by snow crystals. Add to that all the other things sand can abrade and jam up. Desert operations are the harshest environment for aircraft.

Can/how can these be mitigated? For example, I've read accounts of helo pilots in Afghanistan saying that the particle separators of the engines were unable to deal with Afghanistan's "moondust" (sand there can be even 1 micron)

PS's are designed for much larger particles. They just live with low engine life and try to avoid it if possible.  Airliners operating in desert environments tend to have lower engine life as well, as in 50% reductions in life before temperature margins are reached from comp/turb erosion.

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What differences would a helicopter designed to withstand many negative Gs have compared with a normal one?
There are a few helicopters that are able to take negative G's, and to this day only one can actually perform aerobatics (ie. Chuck Aaron's BO-105). The BO 105 is a rigid rotor helicopter that has been modified to perform aerobatics, and sustains loads of +3.0/-1.0 G's.
How would a fully articulated helo be different if designed to withstand negative G's? Maybe -3 or -4 G's. What if the helo is a coaxial counter-rotating one?

I don't think very many pilots can sustain -4 G's.  Negative G's are incredibly uncomfortable & can't really be countered the way that positive G's can. What would be the use case for a helo needing to sustain that kind of negative loading?

Well the limit is usually -3 G's for fighter pilots. On helos I guess just for aerobatics, or maybe some exceptional tolerance in maneuvering

What differences would a helicopter designed to withstand many negative Gs have compared with a normal one? Maybe -3 or -4 G's

Off the top of my head I'd see the following limitations/problems:

G-Force resistance: "Resistance to "negative" or "downward" g, which drives blood to the head, is much lower. This limit is typically in the −2 to −3 g0 (−20 to −29 m/s2) range... Negative g is generally unpleasant and can cause damage. Blood vessels in the eyes or brain may swell or burst under the increased blood pressure, resulting in degraded sight or even blindness".¹
fuel system: more complex and heavy since it has to be able to pump fuel out of the tanks and deliver it to the engines also at negative Gs i.e. when the fuel is floating to the top of the tanks and of the pipes.
oil and cooling system: the same is valid also for these two systems.
structure: all the structural parts which normally work "one-way", i.e. mainly in tension or in compression, now can work both ways and have to be designed accordingly, for example the struts connecting the gearbox with the fuselage.
aeroelastic: the high negative aerodynamic load which has to be created by the blades would bend them downward so much that they would cut through the tailboom and possibly the fuselage; to avoid this a very high mast and/or very stiff blades would be required, increasing weight, drag and vibrations.
aerodynamic: blades should have to be built around an airfoil possessing good aerodynamic characteristics both at very high positive and very high negative AoA; this might perhaps be obtained but only accepting other aerodynamic limitations like higher drag, vibrations and control loads.

¹ https://en.wikipedia.org/wiki/G-force#Vertical

This performance isn't quite that rare. Most heavy combat helicopters (as opposed to lightly-armed recon helos) are also capable of aerobatic maneuvering.
The AH-64 Apache, the Mi-28, and the Eurocopter Tiger are all capable of +3 to +3.5 positive and -0.5 negative g. For the Ka-50, it's also +3.5 g; the negative load isn't stated in any reliable source, but might be in the -0.5 to -1 g range. Coaxials fail catastrophically if their load limits are exceeded, but  perform well up to that limit.
If designing a helicopter specifically for aerobatics, one would want less weight and a smaller rotor diameter. So you'd have many shorter blades, to reduce blade flex. The tail rotor would have to be larger, to provide more control. The reason the Bo 105 only has 4 blades is cost.
A coaxial rotor helicopter would also need even more rotor separation than a combat or utility one, since they bend in opposite directions under load. Coaxials are potentially capable of more maneuvers, but their problem is also design cost and complexity - the transmission and the hub are elaborate and very highly loaded.

So there is really a difference if a helo can pull -1 or -4? I thought there might be some rotor/structural difference

@BanzaiFighterbomber There's not really a need. But the difference would be in how much clearance the rotor has from the fuselage, since it will bend a lot more.


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P-8A Poseidon Airfoil type
Could someone help me find out what airfoil type is used on the P-8A Poseidon symmetrical, conventional, thin or supercritical airfoil?
It seems to be, that the Boeing 737-800, like many commercial jetliners, uses a type of airfoil known as a supercritical airfoil. But I would think the P-8A Poseidon might have a symmetrical one.
Thanks in advance!

Why would you think it might have a symmetrical section?
You need to decide whether you're interested in the airfoils on the wing, the vertical tail, or the horizontal tail.  These are unlikely to be the same.  Also, the surfaces likely use multiple foils.
The vertical tail airfoils are likely symmetrical.
The horizontal tail and wing airfoils are not going to be symmetrical.
There are symmetrical supercritical airfoils.

I know that in Dave Lednicers brilliant website the P-3 is listed with the NACA 0014 and 0012 airfoils. But that does not mean that a symmetrical airfoil is the best choice for a maritime surveillance airplane. Far from it!
The wing of the P-8A is based on that of the Boeing 737-900, and you can be sure that the airfoil is the same. Any change would result in a costly re-certification which the new ex-McDonnell-Douglas management of Boeing will surely avoid like the plague.
The airfoil of the original 737 series was conventional and modified to bring it to a somewhat supercritical shape with the introduction of the classic series (737-300 etc.). Further improvements followed with the redesigned wing of the next generation (737-600 etc.), so that the current 737 models sport a modern supercritical airfoil (which is company-proprietary, of course).


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Obstacle clearance altitude and decision height
can someone explain does OCA that is written in charts coresponds to DA for airlines (its the same altitude), or does airlines choose different DA for example CAT I approach ?


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I have a commercial rotorcraft student and he wants to get rated in an airplane
My new student has his commercial rotorcraft. He is working on his ASEL and he already has most of the experience requirements for PPL. Since he is already commercial rated wouldnt he be taking his commercial check ride in an airplane instead of going for his private?

U.S. and Commercial, his end goal is airlines. He has over 1,000 hours in a helicopter.

He certainly can decide to go straight for the commercial, but he needs to meet the requirements specific to airplanes:

50 hours PIC
10 hours PIC cross country
20 hours training
5 hours of instrument training
10 hours of training in a complex or TAA airplane
day and night 2-hour training cross countries
300 nm "solo" cross country
5 hours night VFR "solo" with 10 landings

Until all of this is done and he passes the commercial checkride, he's limited to barely more than student privileges. In particular, he can't carry passengers or log PIC time while receiving instruction. Some likely sticking points:

he will need to do 5 solo night hours. Since he's not rated in the category, insurance or school policy may be a barrier. Similarly for the long cross country. You can get around this by going along with him as long as you don't actually do any flying or teaching. But then he's essentially paying you to twiddle your thumbs, which eats into the potential cost savings from his point of view and may be pretty boring for you.
50 hours is a quite a bit of time to have to fly solo. Remember that training hours don't count as PIC since he won't be rated in the category (though if he pays you to twiddle your thumbs for 10 hours for insurance purposes, those count).
it's probably cheaper to get the private first. It costs an extra DPE fee but afterwards he can count training hours towards the PIC requirements, instead of those being two separate totals.

So while this is possible, it's probably a better idea for him to get the private as soon as possible and then start working on the commercial.

@RTO Thanks! I knew there was some bit that wasn't quite the same as student privileges, but I couldn't quite remember what part.

And the 50 hours PIC requirement - the 10 hour PIC XC requirement can count towards that as well. I wanted to verify before assuming.

@Miaflyer219 If you mean with you tagging along, then yes, but only up to 10 hours count, and only if you are fulfilling the requirements of 61.129(a)(4), and only if you aren't doing any instruction or flying. If he is solo then any solo XC hours count for both PIC XC and PIC with no limit on hours.

Okay, Thank you


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If I write out a solo endorsement and I decide I want to change the limitations
I am a new CFI and I just gave out my first solo endorsement.
If I write out a student solo endorsement and decide I want to change the limitations. Can I? And if so, do I need to write out a new one or can I cross it out and initial?

This doesn't directly answer your question, which asks about changing an endorsement that you've already written.  However, for future students, you could consider not writing any specific limitations in the endorsement at all, and instead referencing an external document:

I certify that [First name, MI, Last name] has received the required training to qualify for solo flying. I have determined [he or she] meets the applicable requirements of § 61.87(n) and is proficient to make solo flights in [make and model], subject to the limitations specified on [his or her] current Orville's Flyin School Student Pilot Card.

We did this at the flight school I worked at.  Since the Pilot Card (or whatever document you want) isn't part of the student's logbook, you as the CFI can change it at any time, as many times as you need.
(Our FSDO inspector particularly liked that we were able to modify a student's limitations especially with the onset of winter weather.)

I like that thank you


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Is this Common carriage?
It’s been a little while since I’ve studied common carriage vs private carriage.. I need a little advice with this situation. My dad is purchasing a plane (in his name and company) and I’m going to use it to build time but also fly him to places he needs to go for work or if he wants to go on vacation. I also may fly a couple of his employees or friends around at some point. This sounds like common carriage to me but I’m not being compensated in any way. It’s just a gray area..and it feels wrong. Can someone explain how I can do this without breaking any regulations?

Commercial and CFI

Is your dad a rated pilot? If so it's just two pilots flying together.  If not, it sounds like private carriage if he pays all expenses.  It certainly isn't common carriage.  And if you split costs you are just exercising private pilot privileges.  What specifically feels wrong to you?

My dad is not rated. For private carriage but no compensation I guess him paying all the expenses would mean it’s private carriage? And what feels wrong is if I was to carry his friends because I feel like that seems as if I’m willing to carry anyone

Specifically I just wanted to make sure an operating certificate was not necessary.
And for it to be private carriage should there just be a contract in place stating what it is we will be doing?

An occasional personal friend doesn't rise the the level of holding out to the "public".  If this airplane will be in the name of his business it would be wise for insurance and legal reasons to consult an aviation attorney.

What you've described is a corporate flight operation. the majority of these operate under Part 91. So you would be fine as a company 'employee' flying the aircraft for business purposes. For non-business flights, the bigger issue is the IRS. Their rules for use of company aircraft are quite lengthy. You need the advice a good tax lawyer familiar with business aircraft ownership.

Common carriage is a holding out of a willingness to transport people or property from place to place for compensation or hire.
Note that you are being compensated under the FAA's definition since you are logging hours without paying for the plane's operating costs. The FAA construes "compensation" very broadly.
If you are just flying your dad around, you are definitely not holding out and this is definitely private carriage. What's more, you don't have operational control over the plane, so you aren't required to have an air carrier certificate. As a commercial pilot, it is permissible to receive compensation for flying somebody around in their own plane.
Flying around his employees is also fine. It's important that your father, not you, is in operational control. Basically, he provides the plane and the maintenance and such and he just tells you where to fly. If the FAA decides that you have operational control they will consider you to be running an illegal charter.
Flying around his friends is where you get into a grey area. If they are paying you or your father anything then you get into the possibility of being considered an illegal charter operation. This could be considered holding out (and thus common carriage) depending on how strictly your father construes the term "friends."
Disclaimer: I am not a lawyer and you should not rely on me for legal advice.

Ahh okay thank you. It’s all starting to click.
Do I need a contract or something in writing that states what we are doing and how he has operational control?

@Miaflyer219 I'm not sure if that would help or not. Certainly if the contract says that he has operational control but you have de facto control, the contract won't save you. If you do put something in writing it would be wise to consult with a lawyer or run it by your local FSDO to make sure they agree what you are doing is kosher.


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When doing calculations with TSFC of multiple engines, do we add all TSFCs together?
I'm doing an assignment at the moment where I need to evaluate FOMs based off some characteristics of an aircraft. One of the parameters I need to evaluate is the best milage, which I'm doing like so:

For this particular aircraft, the TSFC (c) for each engine is 0.65lb/hr/lb. Do I substitute this into the equation I've attached or do I need to consider the second engine as well? So, would c = 0.65 + 0.65 = 1.3? I'm sorry if this is a stupid question, but everyone seems to be assuming c = 0.65 and I don't understand why when we have two engines.

If both engines burn the same amount of fuel then you need only one number. Obviously that single number will be scaled according to the total thrust and the total flown time.

lb/hr/lb is a hilarious unit. I strongly recommend to use lbf (pound-force) for thrust to avoid confusion. Or even better, use SI units.

The thrust-specific fuel consumption (TSFC) is a measure of how efficient an engine is. It tells you what fuel flow rate is required to generate a certain amount of thrust.
If you have two engines, they will have twice the fuel flow rate, but also generate twice as much thrust. So the ratio does not change:
$$ \text{TSFC} = \frac{\text{fuel flow rate}}{\text{thrust}} \; \rightarrow \; \frac{2 \times \text{fuel flow rate}}{2 \times \text{thrust}} = \text{TSFC} $$
So you can use the same TSFC regardless of how many engines you have.
However, keep in mind that the TSFC of a typical jet engine depends on the engine RPM (N1) and other factors (like speed). Running one engine at maxmimum thrust will not result in the same TSFC as running two engines at half thrust because they will run at a different N1.

Thanks for the answer, that makes sense.


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Hire a Pilot & Rent a plane or need to purchase
I am interested in buying a plane and hiring a pilot on an as needed basis for flights. Is that possible?
Before making the decision to purchase, is it possible to rent a plane and separately hire a pilot? I expect to average 4-5 flight days per month.

Yes, what you are asking about is called private carriage.  If you buy an airplane it's not much different than hiring a chauffeur to drive you in your personal car. The pilot would just need a Commercial Pilot Certificate, and be rated and endorsed for whatever type aircraft you buy. Your insurance company would likely have some additional requirements for total flight hours, time in model, instrument rating, etc.
Regarding the rental part of your question, it would be far easier and better to rent a plane and pilot from the same outfit. Primarily because nobody will rent an airplane to you as a non-pilot.  A lot of places that rent planes also offer private charter service using small GA aircraft.  There's no need or benefit to complicating things for yourself and others by going against an established business model to "separately hire" your own pilot to fly a rental.

The first paragraph is what most corporate flight departments do.  The second paragraph falls under Part 135 charter operations if you want to do more research.


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How does supersonic speed affect vortex lift of a delta wing?
Recent questions have asked why deltas are preferred as commercial passenger supersonic aircraft airfoils.
Sweeping the wings is known to delay shockwave formation as transonic speeds are approached, but what happens to the upper wing vortices of a delta under these conditions?
With straight or swept wings, a "tuck" phenomena is produced by shockwaves forming in different places on the upper and lower wing, resulting in a nose down pitching moment.
The Concorde was known to transfer fuel to adjust to the change of center of lift, but what exactly happens to the airflow when a delta goes supersonic?
Is the "tuck" effect stronger or weaker than a swept or straight wing?

Quite a few topics addressed here, hopefully I can answer everything.
From an historical perspective, the transonic flow has always been considered tough.
Back in the beginning of the transonic/supersonic era (let's say the '40s) the mathematical equations of the aerodynamics had been already known for quite a few centuries. The simplified version of those equations, that is their manually solvable version, quite correctly predict(ed) the aerodynamic behaviour for the subsonic and supersonic (and hypersonic) flows, but they failed to represent the transonic flow where they just gave (give) a solution tending toward infinity. Since computers to resolve the full version of the equations had yet to come, all that was left to do was to come up with a good expression to describe this impossible-to-reach flight state: the sound barrier.
Thanks to the progress both in the computer and in the fligh test field, the sound barrier proved to be not such a destroying effect as predicted by the simplified version of the equations. Anyway even for the today's standards the transonic regime still remains a complicated matter.
The transonic state starts when somewhere on the aircraft a Mach number greater or equal to 1 is reached and this condition is reached even if the whole aircraft has a speed lower than Mach 1 due to local accelerations of the fluid. Tipically this happens on the upper part of the wing, as visible in this picture taken from Wikipedia:

The shockwave which forms could lead to a detachment of the flow behind it with a sudden loss of lift and rise of drag. Anyway modern airfoils for jet airliners are designed to develop and tolerate a mild shockwave, which basically gives no issues.
This event happens also on a delta wing but due to the typical vortices detaching from the leading edge, this phenomenon is a bit more complicated in that 2 sets of shockwaves actually develop.
This picture taken from Schiavetta, Boelens and Fritz (and in turn taken from AIAA Journal, Vol. 35, No. 10, 1997, pp. 1568–1573) depicts the situation quite well:

The first shockwave (left figure, the wing is seen from the back) develops between the upper surface of the wing and the "lower part" of the leading edge vortex and is called embedded cross-flow shock wave. The second shock wave (right figure, dark gray area) develops towards the trailing edge and parallel to it and it is called terminating shock wave.
The behaviour of the aircraft (with or without a delta wing) during the transonic flight or, more generally, during the transition from subsonic to supersonic, is more or less predictable only via complicated and time-consuming simulations and/or flight/wind tunnel tests.
What is generally well known and understood is the overall trend of the aerodynamic characteristics in the subsonic and in the supersonic flow (that is just before and just after the transonic regime): just before, $C_l$ is (more or less) proportional to $2 \pi \alpha$ and the aerodynamic center is at 25% of the airfoil's chord; just after, $C_l$ is (more or less) proportional to $4 \alpha$ and the aerodynamic center is at 50% of the airfoil's chord.
This means that in the transonic regime the lift becomes lower at the same pitch angle and it migrates to the back of the wing. This migration has to be somehow compensated for and in the case of the Concorde it was done pumping fuel in the tail i.e. shifting the CG backward too to make up for the migration of the lift.

Interesting that the migration of Cp is very similar to stall.  This may also help explain supercritical design.  With so much forward speed, one can "surf" the air (Newtonian bottom lift) and minimize drag on top.  Sort of opens the door to low AR supersonic wings, and, lo and behold, we have deltas and trapezoidal designs flying around today (the XS-1 wings were straight).


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What's the most cost-effective way to get a group to get their Junior HPR Level 1 certification?
I'm apart of Civil Air Patrol and trying to get other cadets to get their Advanced Rocketry certificates, but the Challenger (Stage V) phase requires a completion of a NAR Junior HPR L1 certification.
If you want to get this through NAR, you have to acquire a paid membership, that is incredibly costly. I looked into getting the certifications through Tripoli, but they don't offer Junior HPR L1 certifications, only certifications for individuals that are over 18.
Is there any way we can exploit a membership that can be shared by the squadron, or any other ways we can these certifications by the cheapest means?
I will set a flexible budget of ~100 USD if you have any suggestions.
Thank you so much.

I hold a NAR cert level 3. I think it would be best if you contacted NAR directly. I am doubtful that you will be able to join/certify as a group. The NAR provides benefits that cost money (most importantly insurance). It sounds like you want the benefits offered by the organization (certification) but do not want to pay full price. You may want to consider the benefits that your cadets will receive by being a member, including the mentoring and personal relationships that come from the rocketry community. It might even lead some of your cadets into the space profession.

A Tripoli student membership is \$20 per year.
A NAR young senior (21-25) or leader (16-20) membership is \$30 per year.
The cheapest (?) L1 motor reload is an Aerotech H128W which you can currently get for \$24. You can share the casing or borrow it from somebody.
Which would leave you with $40-50 to build the rocket. That's not a lot, but doable.

The rockets and motors are provided and paid for by CAP. I was only basing this off of the cost of getting a certification with no other variables. Thank you for your answer though. Have a great day.


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Flying your own trips part 91 if your jet is on a part 135 certificate
If an individual owns a business jet and places it in an LLC, then in turn places it on a part 135 operator’s certificate, can that individual fly his own trips under part 91?

It's relatively common to see a fractional/charter flight strip (EJA, XSR, LXJ, etc) with the remarks "PART 91 FLIGHT" or "PART 135 FLIGHT." That doesn't change the way I handle the flight as ATC, but it shows that it is possible to operate the same aircraft under different flight rules. Also sometimes you'll see a flight that would normally be "EJA123" be flown as "N123QS" instead, which I imagine is another indicator of 91 instead of 135.

Yes.
FAA operating certificates are applicable to revenue generating flight activity, but there are many reasons besides personal use where an aircraft may be operated under part 91, including flight to a maintenance facility, post maintenance check flights, repositioning for logistical reasons, etc.

I recently had an off-the-record conversation with a highly respected aviation tax attorney about the tax advantages of having your jet on a part 135 certificate. One specific question I asked him was how it affects the tax situation if an owner elects to fly his/her own personal or business flights under part 91. His answer caught me by surprise… He said that unless the aircraft is owned directly by that individual, not by an LLC or other business entity, which almost all business jets are, the aircraft in question cannot be operated under part 91 by the owner. Where is that in the regs?

That sounds like a good question for a "highly respected aviation tax attorney", which I am not.  I am taking my best guess based on some part time work for a 135 operator where we would fly occasional flights under part 91.  I don't know how that may have been affected under different ownership arrangements, and I will delete this answer if it turns out to be wrong...

P.S.  I meant no snark at all in the comment above, just that you should probably seek professional advice for what you are seeking to do.  There may be tax implications outside of the CFRs governing part 135 and 91 operations that will influence your direction.  My point is, to my knowledge there is nothing in the CFRs that directly prohibit this.  The key would be to write it into your operating manual that the FAA reviews and approves.  If it is in there and approved, you would be good to go.


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Is it correct to say that the expected behaviour at a Fly-by/Fly-over waypoint derives from when there was no GPS, but rather just VOR/DMEs?
I was re-looking at Fly-by/Fly-over waypoints:

and discussing with a colleague the topic of how these came about.
Is it correct that this definitions were put in place so that a human pilot could fly between the given waypoints using VOR/DMEs?
This would kind of require having a VOR/DME station under each waypoint where a turn could be expected, so is it plausible?

The concept of a difference between Fly-By and Fly-Over predates RNAV, GPS, and even VOR/DME.
The most basic navigation happens without any navaids at all, and just a map, compass, airspeed, and stopwatch.  Fly to a point defined by some visible feature, reach it, and turn.  That's Fly-Over.
When you draw that route on a chart, in order to be precise you have to account for turn radius to determine the ground track you're flying after rolling out of the turn.  (When I drew low-level navigation charts flying in the Air Force, we had templates that we used to account for the TAS & associated turn radius.)  When the lines on the chart are drawn from point-to-point-to-point, then you have to do something other than flying over the waypoint and then turning to point at the next one... either you lead the turn so that you roll out on that point-to-point straight line course, or else you "S-turn" after flying over the point, to correct back to the that point-to-point line.
When the navigation has limited precision, the difference between the methods doesn't make enough difference to worry about... if my turn radius is a mile, I have 100 miles until my next waypoint, and my wind correction is based on a forecast from a couple hours ago, then the crosstrack error introduced by a 30 or 45 degree turn is probably less than the various other errors that I'll be accounting for.  On the other hand, when I can see my crosstrack error to the 0.01 NM on a modern FMC and Navigation Display, and the autopilot is going to zero that out, it's really nice to NOT have it S-turning back to course.
Most waypoints these days are fly-by so that the charts depict all the airways and tracks as point-to-point, and we let the FMC sort out how far we need to lead the turn, given our True Airspeed right now. (And, of course, the FMC doesn't care if there's a VOR station sitting beneath the point or not.)
But there are times, especially when obstacles come into play, that the procedure designers want the aircraft to stay on the given track all the way to the defined point, and don't start an early turn (in order to prevent something undesirable from happening -- probably, in order to assure clearance from some obstacle or terrain feature).
The difference between the two types of waypoints matters more, now with tighter navigation tolerances and highly precise databases, but the fundamental conceptual difference has been around for a long time, probably almost as long as aircraft navigation & Mr Jeppesen's original charts.
For graphical depictions:

All graphics are mine; I'll edit the text into text as time allows.

@RTO Thanks, I corrected that.  Somewhere back in the memory banks, "fly-past" is another term I've seen used, equivalent to "fly-by", but no reason to confuse things here with two interchangeable  terms.  I agree that RNAV, rather than GPS, is what really raises the distinction between the two types of waypoints, since leading the turn is more a function of an FMS than of just a GPS receiver itself.

@RTO, that ACM discussion was because there was a misunderstanding. DoD and FAA charted waypoints in the plan view the same way, so there was no conflict. The problem is that the 8260.19 is not the authority on charting. Rather, that would be the IAC 4 specification. The current language is more clear, but the result is the same. If the waypoint is used for holding, it is charted  as a fly-by unless it is used in another part of the procedure as a fly-over.

@RTO, and funny enough, I just noticed that I am in the minutes for that ACM issue you linked.

My non-radar instructor once told the class about when broadband radar was installed in Chicago ARTCC. They could see the targets on the scopes, but the system was not yet certified.
Non-radar was based on the "fly-over" concept, so, when a TWA reported over a VOR, but showed he had actually led the turn, they verified his position.
When the controller advised the pilot of the indicated position, the pilot replied "The eye in the sky is upon me".


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Is a yaw string better than an inclinometer in determining coordinated flight in a steep turn?
I feel obligated to go out and try this my self, but maybe someone has done this already and can help.
The issue is that a centered "ball" position is affected by gravity more and more as the bank increases, whereas the yaw string is measuring only airflow in relation to the longitudinal axis of the aircraft.
Another way to determine best coordination is to see if the aircraft holds a level banked turn at a given throttle setting, which, if fully coordinated, would be flying at the bottom of its drag curve for that manuever.  Increase in drag from uncoordinated flight should lead to an altitude loss.
Has anyone seen a difference in the yaw string and ball?  Is it significant enough to differentiate between the two in a steep turn?

Re drag-- google an article called "Circling the Holighaus Way".

String is always better, if you can use it. The ball is less sensitive and can be thrown off a bit, such as when flying a twin single engined, and the rudder is applying lateral thrust to compensate for asymmetric engine thrust.
Coordinated flight in a twin is achieved with a bit of wing down into the live engine, normally 5 degrees, and the result is coordinated flight with the airplane banked and the ball offset in the vial about half a ball to one ball width toward the live engine, because that's where true "down" is.
If you have a yaw string, the string is aligned, which means the airflow is aligned, or it's not.  On any airplane that doesn't have an engine in the way blowing on the windshield, it's desirable to have a yaw string, if only because it's right in your field of view.
Biggest thing is you have to get used to the backwards reference - string to the right requires left rudder for example, opposite the ball indication.  I used to tell glider students, as a memory aid, "Think of the yaw string as a poisonous snake - step away from it."


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Circle just after takoff EWR to EWY
Any idea why this commercial flight (UA1822) would circle as seen in the picture after takeoff at Newark and heading to Key West Int.

Are you looking for a general answer, (or "idea"...) or specific details about this flight?  Because the general answer is probably sequencing delay by ATC.

That looks very much like entering a standard holding pattern... and then exiting the hold before completing one full loop. Possibly a hold over the Colts Neck VOR, but possibly not because the VOR is located basically between the two branches of the Swimming River Reservoir (the body of water immediately west of the text "Long Branch" on the map).

What was its altitude as it made those turns? Level? Descending? Climbing steadily?

velocity 322mph, elevation 11600 feet when the expected holding pattern begun and yes climbing steadily. When coming out of the pattern elevation was closing in on 17,000 feet and velocity 389mph.


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Pitch of a glider with altitude
I'm designing a high-altitude glider. Just to clarify a few points before my question:

The glider will weigh close to 200 grams.
It will be using the A18 airfoil.
The glider is dropped from 80,000 feet.

Please correct me if any of my following statements are incorrect.
I'm calculating the pitch that my glider should be flying at by finding the glide angle. The glide angle does not change with altitude. This means that the pitch my glider should be flying at should not change either, so the elevator should not need to be used, and I will just set my CG to have the nose pitching down at the glide angle. However, the altitude does change the Reynolds number of the air. According to airfoiltools.com, the most efficient alpha changes with Re #, so for maximum efficiency, a glider will need to change the alpha. The most common way I know of changing the alpha is by using the elevator of the glider. This changes the alpha - but it also changes the pitch of the glider! So, the glide angle will change. What should I do about that? Is this the best way of calculating pitch?

Isn’t the alpha the same as the angle of attack, therefore the same as the pitch?  That’s what I’ve always thought. Also, the angle of attack that you’ll need for level flight will change with altitude. The air is thinner at higher altitudes, especially at 80,000 feet. The plane needs more AoA to keep the same lift because there is less air per volume.

Also I would assume because it’d be at such high altitudes, the AoA it would be forced into to retain level flight wouldn’t allow you to compensate for the most efficient AoA from Re number. Dependent on speed though

Pitch =/= AOA @Wyatt

@OrganicMarble Oh interesting. I always assumed they were the same, but no. Good to know!

@RobertDiGiovanni Is there a book you'd recommend?

Fascinated by the aerodynamics aspect of the question, but as a pilot I ought to mention you're going to need a transponder (in my local airspace at least...) and a ton of permissions to drop something from 80k feet with an arbitrary 100NM+wind range. You may want to re-think the 200g payload...

You'll be flying by indicated airspeed so the pitch attitude vs indicated airspeed will be the same at any altitude.

These days much information indeed is available on the internet.  RE: static stability - pitch control is a balance of the wing upforce and the tail downforce torque around the center of gravity.  It's a combination of speed and AoA to produce adequate lift.  Read on.

The angle of the velocity vector relative to the ground reference frame is the flightpath angle, or the climb angle.  Typically a $\theta$ or $\gamma$ depending on which reference you use.
The angle of a reference line on the aircraft to the velocity vector is the angle of attack.  Universally $\alpha$.
The sum of these angles $\theta+\alpha$ is therefore the angle between the reference line on the aircraft to the ground -- this is what an observer on the ground sees as the airplane flies by.  It is sometimes called the 'deck angle'.
At glide, the flight path angle is determined by the lift to drag ratio.  You can draw some velocity diagrams and free body diagrams to convince yourself of this.
$L/D=-1/\tan(\theta)$
As discussed in the other thread, Best L/D will occur at a certain lift coefficient.  Every other lift coefficient will also have a corresponding L/D.  You can calculate the L/D for a given CL from the drag polar.
If you also have a curve for $C_L$ vs $\alpha$, then you can find the angle of attack corresponding to your $C_L$ and also your L/D.
Put it all together and you can calculate whatever you need, $\alpha$, $\theta$, and their sum, the deck angle.

"Pitch" can be confusing in aviation because we have "pitch to the horizon" and "pitch control" with the elevator.
For purposes of clarity let's separate "pitch" into glide ratio and Angle of Attack.
As seen at airfoil tools, the lift to drag ratio àt Re 500,000 will be much greater than 50,000.
Essentially, what this means is you get more lift for the same amount of drag.

In a stabilized glide lift requirement will be constant

You need to study static stability.
Static stability controls the aircraft lift production by controlling airspeed.  No, not with a fancy computer, just by setting the elevator.  Elevator controls angle of attack.
At a higher Re, the wing works better, therefor the indicated airspeed needed for adequate lift (for a given weight) will be less.
As always, when you release your glider, it will dive until (whatever) adequate airspeed is reached to allow it to glide in a straight line towards earth.
The angle of this glide path is determined by have much potential energy (altitude) is consumed per horizontal distance traveled.  This is your glide ratio.
If you have a higher Lift to Drag ratio, Lift being constant, the plane will have less drag to produce the same amount of lift.

thrust = drag

Glider "thrust" is the sine of the angle of its flight path towards earth × weight
What this means is the glider will have a higher glide ratio at altitude.  Again we must realize a glider dropped from 80,000 feet will easily glide more than 100 miles.

for any given elevator setting, the glide ratio will be higher at altitude

For this simple experiment, taken to the extreme of 80,000 feet, plan on greater gliding distance.


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Conundrum about designing a glider relating to coefficient of lift
I'm designing a high-altitude glider. Just to clarify a few points before my question:

The glider will weigh close to 200 grams.
It will be using the A18 airfoil.
The glider is dropped from 80,000 feet.
The pitch should not need to change in flight, it should be equal to the glider ratio.
Because of this, the elevator should not be needed to be used in flight.
Correct me if I'm wrong on the last two points!

I'm doing some CFD analysis in SimScale to calculate the lift coefficient and the drag coefficient to calculate the glide ratio of my glider. SimScale requires a speed for the glider, which makes a lot of sense. This is all good because the coefficient of drag and lift shouldn't change with different speeds, so I don't need to know the speed beforehand. That being said, I understand that you can commonly find the speed that your glider should be flying at by multiplying the stall speed by 1.3. However, something that does change the coefficient of lift is the AoA, which is determined from Reynolds number which is derived from number of things, including speed. So, to calculate the coefficient of lift I need to know the AoA, but to calculate the AoA I need to know the coefficient of lift to calculate the stall speed.
How do I get around this?
I'm probably missing something, please excuse my ignorance because I'm new here!

This is all hypothetical, yes?  Or is somebody actually taking a 200 gm (7 oz) model airplane up to 80,000' to drop it into imprecisely known winds, almost certainly never to be seen again?  While certainly possible, that seems a bit expensive, and to what purpose?  So is this intended to actually fly, or just design it & that's that?

Highly related: https://aviation.stackexchange.com/q/104486/7532  and https://aviation.stackexchange.com/q/104580/7532  It's not clear how this question is materially different from these others.

@RalphJ This is certainly happening. No, it isn't very expensive either. Everything is under a thousand dollars.

@RalphJ Re: Highly related. Yes, I understand this concern. I've just started learning aero so those posts bared lots of ignorance. I've also split my questions into different posts.

The change in Reynolds will change the (stabilized or trimmed) glide airspeed without affecting the AoA, correct?

Before putting something up to 80 000 feet in the hope of flying it all the way down, I would start with removing all the zeros and work your way up from there. Build the thing, get it flying from 8ft. If that goes well, next step is to master 80ft. Then 800ft, 8000ft and finally FL800. Somewhere between 8 and 80ft you'll figure out the trim setting, without CFD, AoA and Reynolds

@DeltaLima That is a good idea, but I would like to figure out everything with math to find the most efficient way possible to do this.

To start with, you can ignore changes due to compressibility (Mach number) and viscosity (Reynolds number).
Your aircraft will have a drag polar approximately of the form:
$C_D=C_{D,0}+K\,{C_L}^2$
You can do your aerodynamic analysis to determine the $C_{D,0}$ and $K$ coefficients.
This form of drag polar achieves best L/D when $C_{D,0}$ = $K\,{C_L}^2$.  You should be able to derive this, it is fun.  For now, you can just trust me.
From that, you can solve for the lift coefficient at best L/D.  This will be:
${C_L}^*= \sqrt{\frac{C_{D,0}}{K}}$
This tells you the lift coefficient you must fly at to achieve best L/D.  This corresponds to some angle of attack, but I don't care about the numeric value of angle of attack just yet.
In steady flight, the lift coefficient of an airplane is:
$C_L=\frac{W\,cos(\theta)}{q\,S_{ref}}$
Your glide slope will be shallow, so we can make the small angle approximation $cos(\theta)=1$.  If you want, we can revisit this later.
$C_L=\frac{W}{q\,S_{ref}}$
Here, $q$ is the dynamic pressure $q=0.5*\rho*V^2$, $W$ is the weight, and $S_{ref}$ is the reference area of the wing.
For a given wing loading ($W/S_{ref}$), your aircraft will fly at best L/D at a specific value of dynamic pressure $q$.
This $q$ corresponds to a single value of equivalent airspeed.  Or, if you want to work in terms of true airspeed, the TAS will vary with altitude.
I suggest you pick a typical altitude to do you calculations around.  Later, you can come back and check how much changing altitude changes things.
(For the record, changing Mach and Reynolds number (which are altitude effects) will change the $C_{D,0}$ and $K$ coefficients slightly.)
OK, so now we know that we want to match a desired ${C_L}^*$ with a speed and altitude in the form of dynamic pressure $q$ and wing loading $W/S_{ref}$.
While we are designing the aircraft, we have control of $S_{ref}$ in order to get the desired cruise $q$ to match ${C_L}^*$.
Once the aircraft is designed and built, we can't change wing loading any more (except by adding ballast).  So, then we must choose $q$ to get the desired ${C_L}^*$.
Now we'll assume you have all this worked out -- your wing is sized ($S_{ref}$) to make your aircraft $W$, $C_{D,0}$ and $K$ reach best $L/D$ at the desired dynamic pressure $q$.
The next step is to trim the aircraft to actually fly at the desired $C_L$.  For this you need a horizontal tail / elevator or control of the CG.
Since you want to fly at the same $C_L$ for the entire flight, you won't need to change this in-flight (ignoring Mach and Reynolds number effects), but you do need to figure out how to get the airplane to trim where you want it.
This is another question, so ask for more if you need it.
The short version is to do a simple linear trim problem to achieve the desired $C_L$ at $C_M=0$
Edit:
Summary:

The drag polar of the aircraft ($C_{D,0}$, $K$), has a single ${C_L}^*$ where it achieves best L/D.

Start by estimating the drag polar for an assumed wing area / design.

Once you know the polar, ${C_L}^*$ is easy to calculate.

The flight condition can be thought of as a combination of Velocity $V_{TAS}$ and density $\rho$ (two quantities) or as the dynamic pressure $q$ (one quantity) or equivalently $V_{EAS}$.  I will use $q$.

The $C_L$ that you fly at is determined by the weight $W$, the wing area $S_{ref}$ (combined into the wing loading $W/S_{ref}$), and the flight condition $q$.

As a designer, I can have a target flight condition $q$ and I can control the wing loading $W/S_{ref}$ to make that flight condition align with ${C_L}^*$.

Once the drag polar and wing loading are determined, I do not have choice in flight condition -- there is a single flight condition where the aircraft will achieve ${C_L}^*$.  We're stuck with what you get there.

Even after all this, you must also trim the aircraft such that $C_M=0$ at the desired ${C_L}^*$.  Trim can be achieved by deflecting control surfaces, moving the wing, moving the CG, etc.

Great post, thank you for going in depth. I'm a bit confused though; what does the outcome of all this mean?

"Start by estimating the drag polar for an assumed wing area / design." - How does one do that? (Thanks for the edit, BTW).

You could run your CFD code at three angles of attack and fit a parabola to CD vs. CL and a line to CL vs. Alpha.  Everything I said above also applies for a polar of the form $C_D=C_{D,0}+K_1,C_L+K_2,{C_L}^2$ -- which is more obviously a parabola and takes three points.  While you're at it, fit a line to $C_L$ vs $\alpha$ and $C_M$ vs $C_L$, you'll need those later.  Pick moderate $\alpha$, stay in the linear range of lift, avoid stall.  There are less computationally expensive ways, but if you're set up with CFD, this way is conceptually simple.

However, something that does change the coefficient of lift is the AoA, which is determined from Reynolds number.

The AoA is determined solely by the relative angle between the lifting surface and the airflow, it's a pure geometric matter (picture from Wikipedia):

The Reynolds number has only an impact on the aerodynamic coefficients. For example the following picture shows the drag coefficient for a sphere as a function of the Reynolds number (plot source):

I don't know at which Reynolds numbers your glider is going to fly, but within a more or less ample range of Reynolds numbers the changes in the coefficients can be neglected.

So, to calculate the coefficient of lift I need to know the AoA, but to calculate the AoA I need to know the coefficient of lift to calculate the stall speed.

The process is much easier: according to the flight condition you want to achieve, you determine the needed lift and lift coefficient and the relevant AoA as a consequence.

"The AoA is determined solely by the relative angle between the lifting surface and the airflow, it's a pure geometric matter"
So how do you explain this airfoil plotter, where the AoA with the best Cl/Cd changes with Reynolds number: http://airfoiltools.com/airfoil/details?airfoil=a18-il?

As said in the answer, the aerodynamic coefficients change with Re and as a consequence their ratio as well, that's why $C_l/C_d$ changes.with Re.

I would start with Reynolds number and evaluate your anticipated speed envelope.
The wing you have chosen is good for low Re number aircraft.  Why not use the same airfoil for you horizontal stabilator?
Re number = (chord meters × velocity meters/second ÷ kinematic viscosity 1.48 meters$^2$/second) × 10$^5$
A 200g glider with a chord of 0.14 m at sea level will glide at around 5 meters/second, giving a Re value of around 50000.
Using the True Air Speed Calculator we find TAS at 80,000 feet will be around 35 meters/second, for a Re value of around 350,000.
All other airfoiltools graphs can be discarded.
The increase in coefficient of lift at the higher Re number will result in a slower glide speed.  As lift is proportional to V$^2$, the change in airspeed will be the square root of the change in coefficient of lift.
When a similar analysis is done for your stabilator airfoil, you may find that trim changes in flight may not be necessary.
Find you gliders Vbg at sea level (or where you live).  To be safe, trim it to a slightly higher speed (lower AoA) for your first flight.
Based on your glide ratio and prevailing winds, your glider may easily travel over 100 miles.  It might be wise to add some rudder and ailerons to let it spiral down, yet another reason to go with a higher (safer but less efficient) glide speed.


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Where do i find data about the Kamov Ka 50?
I am a student in Aerospatial Engineering and i have to make a project about this helicopter. Professor asked for some data like: Rotor blade airfoil, Rotor blade width at the beginning and tip of it, at what height can it safely land just with autorotation, and so on. I have no idea where to get this data from, can someone help?

Hi and welcome to Aviation. Unfortunately resource location questions are off topic here.

I can pretty much guarantee, that data about the exact airfoil or dimensions etc. is not publicly available but rather classified information, as it would allow to calculate classified performance data of this military helicopter.
That being said, you can make good guestimates from photos which you can find on the internet. Especially root and tip chord and similar data can be extracted from photos which also show known dimensions. Perhaps you can also look at some photos of the blades and "guestimate" the airfoil. You could also assume that this helicopter has the same airfoils as other helicopters. For example, the Bölkow Bo-105 has NACA 23012 airfoils, so perhaps it is safe to assume that the Kamov KA 50 has this or a similar airfoil? I would bet that there is not much difference.


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Are ICAO airport codes starting with X-ray valid?
The Wikipedia page on ICAO airport codes currently states that

ICAO airport codes do not begin with I or J or X or Q

and later

In Russia, Latin letter X, or its Morse/Baudot Cyrillic equivalent Ь, are used to designate government, military, and experimental aviation airfields in internal airfield codes similar in structure and purpose to ICAO codes but not used internationally.[2]

However to my understanding, while the link does show that military bases use code starting with X-ray, it does not prove that these code are considered invalids by the ICAO.
On the other hand ourairports database consider these valid ICAO codes.
Now given both are crowd-based and without a link to an official source it is hard to know for sure which is correct. Is there some official document clarifying the matter?
Unfortunately the official ICAO airport database is very expensive to get ($1,500).

My question is, are these ICAO airport codes, or not? And where's the source for the yes/no answer.

In short:
There are no ICAO locations beginning with X, as evidenced by the official list in Doc 7910.
ICAO location indicators are used to route messages between stations of the aeronautical fixed service (AFS), the service collectively provided by all ICAO members to facilitate international aviation.
Doc 7910 contains the list of all international location indicators, and each State also includes the indicators it manages in its AIP, at section GEN 2.4 (example).
Note ICAO 4-letter indicators are not always associated with a particular aerodrome, e.g. France uses the common address LFZZ to receive and transmit NOTAMs.

Location indicators are AFTN locations.
Civil aviation entities across the world rely on the aeronautical fixed service (AFS) for the exchange of messages and data to coordinate international civil air traffic. They are connected by the aeronautical fixed telecommunication network (AFTN), a kind of teletype network.
Each entity participating to the AFS receives a 4-letter location indicator for message routing through the AFTN. Large aerodromes, potentially involved in international traffic are part of the AFS, Aerodromes not involved in the AFS are not connected to the AFTN, they don't need an ICAO indicator.
The 4-letter indicator is then suffixed to get unique addresses for all services (tower, weather services, control center, air authorities...).
Routing scheme
The first letter of a location indicator delimits routing regions:

(From Doc 7910)
Within a region, a second letter is used to create blocks allocated to States and Territories. Many States need blocks in different routing areas (e.g. France uses blocks LF-Metropolitan France, TF-Antilles, SO-Guiana, FA-Polynesia and a few other).
The third letter is a communication center used by a group of stations, and the fourth letter is the station.
Blocks are allocated by ICAO, the last two characters of the location indicator are determined by the State/Territory to which the block has been allocated.
Three more letters are added to the location identifier (according to Doc 8585 - Designators for Aircraft Operating Agencies,  Aeronautical Authorities and Service) to reach a particular service. For examples:

YXY to reach a military organization
ZTZ to reach the control tower
ZZZ to reach an airborne aircraft

A message sent to EBLGZTZX is sent to EBLG = Liège Airport, ZTZ = control tower. X is a filler character for a fixed length of 8 characters.
Indicators vs. Aerodromes
A 4-letter location doesn't necessarily designate an aerodrome, e.g., these locations are not associated to a particular aerodrome: LFFF (Paris UAC control), LFYX (Air Force Paris area), LFZZ (France NOTAM management).
Many aerodromes have no ICAO location indicator, because they are not involved in international operations.
Some location indicators correspond to AFTN unconnected stations.
No X area
There is no routing X area, and as a matter of fact:

The last W block allocated by ICAO is WS, to Singapore.
The next block is YA, allocated to Australia.

Often cited, XBRO is not an ICAO indicator for Broadford_Airfield. However many databases include(d) it, e.g. OurAirports (note the code in the URL), and flight simulators have reused it, this creates problems.
Any organization can create 4-letter identifiers, and identifiers starting with X can exist. But they have no link with AFS/AFTN, they don't represent known ICAO locations.
I don't know why X was excluded as an area (contrary to I and Q for obvious reasons). Perhaps this is linked to the fact X is used as a filler, or could be used to strikeout a whole location designator.

Fantastic answer, thanks!
Since you're also mentioning OurAirports, if one's is using their database, they should absolutely use gps_code field for ICAO code, and  not ident, because despite being in 99% of the time the same, it sometimes break when the airport was added with the wrong code (ident being immutable).  More information in their database dictionary


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"Air Force One" outside of US airspace
Does the USAF aircraft carrying POTUS use the call sign "Air Force One" when in other countries' airspace? I could see some countries getting a bit miffed, seeing as they have their own air forces, and might ask that POTUS' aircraft use the call sign "US Air Force One".
Or is there an international agreement that POTUS is Air Force One when in other airspace?

I guess such military flight cannot be considered international civil aviation as defined by Chicago Convention and therefore cannot benefit from the related treaties. It likely requires a diplomatic clearance to operate in other national airspaces. The callsign is likely part of the clearance, as well as weapons aboard and non-ADS-B use, etc. This would be the same for foreign diplomatic missions in the US.

Related: Do military aircraft overflying other countries' airspace require special clearance?

The real question would be "Is "air force" a callsign reserved for the USA with ICAO worldwide or only domestically with the FAA?" Really it comes down to what is filed in the flight plan.

Other countries have other callsigns for their airforce. Denmark has "Danish Airforce", Bulgaria has "Bulaf", Finland has "Finnforce", Singapore has "Singa", Romania has "Romaf" etc. etc.

Great question! The answer to that is the "Air Force One" callsign is reserved for planes carrying the US President WHEREVER they fly, whether it be to an ally state like France or a far-away distant regime like Myanmar.
It is true that other states also have their own "callsign" for planes carrying their heads of State/Government. But it will not use the "Air Force One" callsign. For example, Canada calls their jets carrying their Prime Minister "Can Force One", France calls it "Cotam One", and Japan calls it "Japanese Air Force One" (it's a long-ass name but they had to call the whole callsign!), etc. Conclusively, I am not sure if it is an international agreement or how the US managed to keep the callsign very exclusive to themselves and enforce it in foreign airspace, but you will never find a second "Air Force One" callsign.
Hope this helps!

"the "Air Force One" callsign is reserved for planes carrying the US President WHEREVER they fly"...... reserved with whom, under what authority?

@user95861 hi! As I said, I do not know how they do that, but not even the Chinese or Russians have been recorded to use this callsign. For some reason, the US Air Force has been able to keep the callsign exclusive. Don't ask me how... all I know is that I observe and see that no other countries have used "Air Force One" callsign but only have the equivalent.


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How come there is no attempt at an "inter-meshing rotorprop" high-speed helicopter?
There are some prototypes of high-speed helicopters, such as the Eurocopter X³ (X-Cubed), Sykorsky X2, and older ones such as the AH-56 Cheyenne.
At high speeds, forward and backward traveling rotor blades relative to high speed flight path induce asymmetric lift and drag: Forward helicopter flight speed minus backward rotating rotor blade speed may equal zero relative speed, or even a negative value.
So, how come there's no [known] attempt at making an inter-meshing rotorprop helicopter?

Inter-meshing like K-MAX
Rotorprop (pusher) like Sikorsky S-66

It seems to offer a far more symmetric configuration than the examples mentioned above, and as illustrated below:

Below an illustration of assymetry at some positive AoA (view is normal to relative wind):

Beyond the obvious (Kaman is likely not as rich in R&D funding as the others) what about the intermeshing rotor configuration in Kmax makes you belief that it's a better fit for high speed flight?  Your diagram only tells part of the story.

It really doesn't matter the two rotors are coax or inter-meshing if rotor is not rigid: they are both as not suitable for high speed as conventional.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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Truss-braced wing and variable geometry
Reading about variable geometries, swing wings, rearward or forward sweep, oblique wing designs pro and cons, and Truss-braced wing designs, are there studies or experiments involving truss-braced wings and variable geometry?
One of the main disadvantage of oblique wing being maybe the pitch-roll coupling, and one of the main disadvantage of swing wing being displacement of center of lift with relation to center of gravity, using structural properties of truss-braced wing by having pivot points on each half-wing located somewhere at each half-wing individual center of lift might mitigate those issues. (probably adding new issues too)

@RobertDiGiovanni im not sure i understand. If the strut slides at the bottom, it's another variable geometry hinged at wing root like F14. Here instead the strut is static, and wing root somewhat slides in upper fuselage.

It's a matter of geometry.  Scale up the drawing and see what happens to the length of the mount points.  It would be possible to slide the wing mounts forward and have the truss pivot on either side.


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Timing not authorized for defining the MAPt
Does anyone know why some charts have written "Timing not authorized for defining the MAPt"?
Could this happen when the DME required is written on the chart?
Thank you,

Can you post an example approach procedure with this notation?  Thanks -- and welcome to Av. SE!

“DME required” typically means that DME is required somewhere from the IAF through the missed approach. It does not mean that DME must be used to define the MAP. Some nations, following PANS OPS, will always show a rate of descent table that includes distances/timing from the FAF to threshold. The timing note indicates that the table cannot be used for defining the MAP.

I can share two examples: EFOU VOR RWY 12 (https://www.ais.fi/eaip/002-2024_2024_04_18/documents/Root_WePub/ANSFI/Charts/AD/EFOU/EF_AD_2_EFOU_12_VOR.pdf) and EFOU RWY 30 (https://www.ais.fi/eaip/002-2024_2024_04_18/documents/Root_WePub/ANSFI/Charts/AD/EFOU/EF_AD_2_EFOU_30_VOR.pdf)

From the designer's point of view, using PANS-OPS criteria will be necessary to do a study using Appendix to Chapter 6 - Refined Method for calculating MAPt and transitional tolerances for a Missed Approach point defined by a distance from the FAF?
Thank you,

There are two cases for this mention to be added:

The MAPt is collocated with a facility or a fix, there is no reason to use a timing from the FAF to identify the MAPt. This case is the one justifying the mention in the two | examples you have linked.

The MAPt is identified by a facility, but is not collocated with this facility. The timing may be useful to fly the final approach, however there can be an obstacle in the missed approach segment or some other reason (e.g. the proximity of a restricted area or a noise abatement procedure) which requires the missed approach to be executed accurately. In this case, if identifying the MAPt by timing is deemed not safe enough, the mention will be added (example below for Tokyo Intl).

From ICAO PAN-OPS vol 1 (Doc 8168):

8.4.4.2 If the missed approach point (MAPt) is defined by a facility or fix at the MAPt, the procedure will be annotated “Timing not authorized for defining the MAPt”.

6.1.5 When the MAPt is defined by a navigation facility or a fix, the distance from the FAF to the MAPt is normally published as well, and may be used for timing to the MAPt. In all cases where timing may not be used, the procedure is
annotated “timing not authorized for defining the MAPt”.

Example with such mention, LOC W RWY34R at Tokyo Intl. The MAPt is defined by the distance 0.7 NM to the ILS DME (ITC):

Okay, if timing from the FAF to the MAP isn't published, you can't use timing -- that's fair.  But, what sorts of characteristics of an approach cause the designers to decide that timing isn't an appropriate way to identify the MAP (in turn leading them to not put the timing info on the published product)?

@RalphJ: Because there is an obstacle in the missed approach segment, it is not safe to use timing for defining MAPt.

Could you share this approach with us? What is the airport?

@SarahFoxes: I added a link to this plate.

Per TERPS 2-9-8(c), this can happen when there is excessive fix error on the PFAF (precise final approach fix) and there are obstacles on the missed approach segment. The fix error on the PFAF is not allowed to exceed 1 nautical mile if timing from the FAF is used to identify the MAP, unless an additional buffer area is provided before the missed approach surface begins to account for the excessive error.

Is this based on timing inherently having greater error than DME (believable) or RNAV (believable) or NDB station passage (maybe) or other methods?  How is it determined that a given fix has however much error to it?

@RalphJ When the PFAF is identified from some ground station, the type of station, the distance to the station, and the bearing to the station all affect the fix error. DMEs has some ± distance or % of distance associated with them and VORs have some ± angle associated with them. TERPS has a bunch of pictures showing this. The issue is not just that timing has some inherent error but that if you are timing from the FAF which itself has a large error, then you have to add these errors together. If there is a way to identify the MAP independently then the error on the PFAF doesn't matter.


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Estimate the drag caused by wind on a free falling body
I'm trying to solve the following problem for my masters thesis: Estimate the drag caused by wind on a free falling body from H altitude. H is in range of 300-1000 meters.
What would be the best way to calculate the drag caused by the wind ?
Assuming I have wind estimates for each altitdue.
Assuming I have drag coefficents of the body for 0.01-0.99 mach numbers.
Assuming I have duartion of the free fall.
Assuming I have center of pressure and center of gravity.

You simply have to write the equation of motion adding the drag as an opposing force of magnitude $D=½ \rho V² S C_D$ with $C_D=C_D(Re, Ma)$ and $V=\frac{dx}{dt}=\dot{x}$. If you post your try we might be able to find the error

This might turn out to be extremely complicated.  First of all, does the problem start with the body at 0 velocity relative to the ground, or at terminal velocity (instantaneous acceleration of zero), or at some other condition?

Drag caused by wind. Ha!  This would involve drift (horizontal) from drop point.  But it may be solved in a similar manner, either as a vector of relative wind or as the wind vector alone.  If you truly want to go Masters level, you may wish to study propeller disk drag on an air craft.  Is it just the drag equation, or does the accelerated airflow around the prop (airfoil) somehow deflect the oncoming air (from aircraft velocity)?  How does this compare to the drag of a prop that isn't turning?

One complicated bit is that any horizontal wind relative to the object (i.e. any time the object's instantaneous horizontal groundspeed is different from the wind speed at that altitude) will also change the vertical force oposing gravity, even in a situation where the only aerodynamic force components are drag components, not lift components.  (E.g. a non-spinning round ball, or a bomb with fins large enough (in relation to angular inertia) to keep it perfectly "weathervaned" to point directly into the instantaneous relative wind at all times.)

To understand this, visit this Physics SE question, paying special attention to the first paragraph below the itemized list of three items:  https://physics.stackexchange.com/q/491600/236723.  Now add wind shear into the mix (different wind speeds at different altitudes), and it seems that you will have a very complex problem.

An excellent reference, with an interesting result (because of V$^2$).  We may have Thesis yet, but first ... to the wind tunnel for some data!


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What is unarmed weight?
 I'm doing research about fighter planes, and I've noticed two terms: unarmed weight and empty weight. I've tried to find what the difference between these two terms is, but I can't find a definition for unarmed weight. Is it just the weight of the fuel and no weapons (hence, unarmed), or is there a more specific definition connected to some other weight measurement, because there are several of those and I wouldn't know exactly what unarmed weight is connected to.

Empty is no fuel and no munitions.
Unarmed is with fuel and no munitions.
Where are you seeing these terms?

I'm seeing them on wikipedia pages and a couple other sites, specifically in reference to the Mig-25, with a difference of 29k kg for unarmed weight and 20k kg for empty weight. Wanted to confirm the difference between them.

It would be best, @Chickendragon, to put this info (including the links to wiki) in your question.


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Odd aircraft lights. 3 long warm white/yellow rectangle lights- solid, not blinking
Lights seen in busy commercial air traffic zone over Essex County MA at 7-7:15pm in a North/NW heading, approximately 320-330degrees. Unlike the other traffic,this was completely silent from the ground and seemed to be gliding at an angle, which leads me to believe it may have been a military delta wing aircraft with us only having a view of one side from our perspective(V shaped).
What I find particularly odd is that they didn’t have any Red & Green nav lights, no spotlight, no landing lights. In a crowded area that would seem unacceptable for an  Air Force training flight. I checked the air traffic and none of the commercial airliners matched the heading at the time.
So is this something any of you have seen or anything you have knowledge of? Someone suggested it might be a test flight of the new B-21 but as far as I know that’s only happening in CA at the moment. I doubt they are doing continental test flights already, but maybe?
No, I don’t think it was a UFO before anyone chimes in. Thank you for any help!

Keep in mind that quadcopter-style "drones" are now ubiquitous and may be showing any random display of lights.  The days of anyone paying much attention to reports of weird lights at night are probably pretty much over.  Even more traditional winged rc airplanes are also venturing out night with all sorts of bizarre lighting patterns.  PS if this is a report from a cockpit and you believe that you were too high for such things to have been a likely explanation, you might want to clarify that this is a report from an airplane not from the ground, and also mention your altitude.

But I've flown a radio-controlled model glider via conventional line-of-sight flying (no camera goggles etc) up to 4K AGL in daylight, and a well-lighted object of similar size obviously could be seen much further away at night.  Not to mention all the possibilities for semi-autonomous or fully-autonomous flight for such vehicles these days.

"I don’t think it was a UFO" - Yet you came here to ask about an object that was flying that you couldn't identify...  ;)


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What are the Cessna 172 carburetor icing chart data points?

I am looking for the polynomial equations (functions)
for each of the curved lines.
I have been looking online and I am unable to find the formulas and the data points for this chart (Cessna 172).
I am trying to program an application that can calculate the icing probability for a Cessna 172. By the way if you have the data points for other types of engines and planes that would also help me a lot!
What are the Cessna 172 carburetor icing chart data points?

As far as I'm aware, that chart doesn't vary by aircraft.

@Jamiec no clue. I did do some research on the engine that Cessna's 172 use and it is the Lycoming O-320 Series. Skyhawks and other types of airplanes use it too. I went intot the engine documentation and I haven't found anything on carburetor icing graph. This is what I found so far: https://github.com/jnbdz/aviation-quickstarts/tree/main/Airplanes/Cessna/172/Engines

What's the source of the figure you are asking about?

@Adam good question. I have no clue. Even ChatGPT seems to be unable to find that information.

@mins thank you so much! My googling skills are terrible! Once I have a minute I will explore it in more details.

This chart is not related to any particular aircraft. It was initially published by the Chief Accident Investigation Establishment in Canada:

Source.
But it has many variants and yours doesn't match exactly the original.
Still the curves can be easily approximated as polynomials using appropriate fitting functions. For example sampling the curves at a few points, using a piece of Python code and Numpy polyfit (least squares and Vandermonde matrix), I got polynomials matching pretty accurately the chart.

The corresponding coefficients for polynomials (up to degree 5), in order of increasing power, from the outer curve to the inner one:
[19.634  1.688 -0.044  0.    -0.    -0.   ]
[ 1.5896e+01  1.9160e+00 -1.4600e-01  1.1000e-02 -0.0000e+00  0.0000e+00]
[ 8.797e+00  2.878e+00 -3.080e-01  2.400e-02 -1.000e-03  0.000e+00]
[ 4.955e+00  3.183e+00 -5.500e-01  6.500e-02 -4.000e-03  0.000e+00]

The code used:
for a in [c1,c2,c3,c4]:

    # Unpack input array
    dp, t = a[:,0], a[:,1]

    # Coeffs - Low to high power
    c = npp.polyfit(dp, t, deg=5)
    print(np.round(c, 3))

    # Curve samples
    dp_s = np.linspace(dp[0], dp[-1], 20)
    t_s = npp.polyval(dp_s, c)

    # Plot everything
    ax.plot(t_s, dp_s, c='m', lw=3) # samples
    ax.scatter(t, dp, c='m', s=90) # input points

Arrays c1, ..., c4 contain the sample values with dew point in first column and temperature in second column:
c1 = np.array([[-13.9, -13.7],
               [-9.2,  -0.],
               ...

Since the original functions are not injective (folded curves), the polynomials are the inverse functions, they take the dew point as the variable, and return the temperature, $t = \mathrm p \,(dp)$.


text:
Does the lift force comes from the bottom or from the top of the wing?
My team and I discussed yesterday how lift is generated based on the pressure analysis. The most popular explanation from books and other sources is that pressure is lower on the top and slightly greater on the bottom and "pressure difference" makes the force. My argument is that because the pressure on the top is lower than the atmospheric pressure the pressure on the bottom makes a pushing force upwards. Others from my team say that because the pressure on the top is lower there is a great pulling on the top and a small pushing force from the bottom. The thing is that our simulation program can generate this kind of plot:

And its not only our simulation software, this type of image can easily be found all around the internet representing lift. I say this lines are just a measure of pressure, the orientation shows if its lower or greater than the atmospheric pressure and the direction is normal to the surface just because it is convenient. Some of my teammates say that these vectors are actually force vectors that simulate the lift force direction if we sum all of them. What do you think?

It does not. I asked for a "push" or "pull" force. It's intuitive I'm certain you got what I mean. The details about pressure or Newton's laws we already know.

What about this one?

Vacuum (lower pressure) can certainly create a "pulling" force.  The ratio of "pushing" from the bottom and "pulling" from the top is strongly dependent on Reynolds number.  But one can see how some might see it as all "push" from higher to lower pressure.

There cannot be any "pulling" on the top. There is only reduced pressure.

In short: Both the pressure on the top side and on the bottom side push against the airfoil. The airfoil is overall pushed up, because the force on the bottom is larger than the force on the top.
Lift results from the pressure difference. Still we could say the top side (usually) participates more to the creation of this difference than the bottom side, in the sense the top side decreases the pressure below the ambient pressure more than the bottom side increases the pressure over the ambient pressure.

There is no "physical" lift force, lift and drag are engineering concepts. Lift is the vertical portion of the sum of all tiny pressures acting on the airfoil surface.
Conventional representation of the pressure variation created by airflow on an airfoil with positive lift:

Source.
Be aware of two confusing aspects in the picture:

There is no volumetric notion here , the lines indicate the pressure magnitude and direction directly at the airfoil surface. Pressure at any distance from the surface is unknown.

Arrows seem to indicate a suction effect on the top side, and a pushing effect on the bottom side. Remember pressure only pushes against the surface, it doesn't pull. Arrow indicate the sign of the pressure differential with atmospheric pressure, not the direction of the force. Arrows are exactly redundant with color here.

Red lines depict how much the pressure is in excess of the atmospheric pressure, blue lines depict how much the pressure is below atmospheric pressure. Pressure acts perpendicularly to the airfoil surface, hence the direction of the lines.

The bottom side sees a local pressure gradient which values are greater than atmospheric pressure, hence overall this side tends to push the wing up, more than atmospheric pressure alone.

The up side (unfortunately called the suction side in English) sees a local pressure gradient which values are lower than atmospheric pressure, hence overall this side tends to push the wing down, but less than atmospheric pressure alone.

Because the top pressure gradient is overall smaller than the bottom pressure gradient, the latter wins, the wing moves up. Whatever, all these pressures with disparate magnitudes and orientation don't combine in a perfect vertical force, the force is actually oblique. (And because the pressure differential at the leading edge is not the same than at the trailing edge, some positive or negative pitching moment is created too.)
Pressure force and aerodynamic force
To simplify, engineers mathematically integrate tiny pressures into a single symbolic larger force, the total aerodynamic force. For the airfoil above, the total force has an oblique orientation, up and back, hence the airfoil is pushed up and back.

Source.
Breakdown of the aerodynamic force into lift and drag
Engineers like to split this total force into a sum of two other symbolic forces of particular directions, drag and lift, to ease common computations. Note this decomposition is purely mathematical and has no actual physical interpretation.
The handy directions are the vertical and the direction of the flow. Lift, the vertical component, can be compared to weight to determine whether the aircraft will gain or lose altitude. Drag, the flow direction, can be compared with the propulsive force, to determine the velocity of the aircraft.
Varying the total aerodynamic force
The shape of the airfoil is the mean to change the magnitude and the orientation of these tiny forces, so every airfoil has a different distribution of the tiny pressures, and the distribution can change with the orientation of the airfoil:

Source.
Ultimately lift and drag conventional forces can be adjusted. With a negative angle of attack (top case), the aerodynamic force can be oriented down, lift has a negative magnitude.
From pressure to speed
Note how the pressure on the top side is mostly smaller than atmospheric pressure (- sign) and larger on the front part of the bottom side. Intuitively if you oppose a fluid (air or water) flow with your hand, or if your hand moves in a stationary fluid, you sense a force, fluid pressure has increased. The reason is because the flow has been slowed down. Conversely, if the flow is accelerated by some mean, pressure decreases (hand is not a good sensor for this negative pressure change).
Thus the sign of the pressure on the top side (also) indicates where the flow accelerates and where it slows down. Namely it always accelerates after some distance from the leading edge, even when the angle of attack is negative. This acceleration, combined with the deceleration on the bottom side results in a up force. Airfoil designers look for the "best" combination of changes to the flow speed at different angles of attack, e.g. one resulting in a near vertical sum.
Which side creates more pressure?
The total lift is the integral of all pressures, hence it corresponds to the difference between the sum of vertical components of all red and blue arrows of the first picture. Without performing the actual integration, it appears on the examples provided, the largest part of the lift comes from the top side, specially at high angle of attack. The stall angle is the limit, past the stall angle air acceleration starts to be problematic due to the apparition of vortexes. This is the direct answer to your question.

Pressure always acts normal to a surface.  It isn't just convienent.
Away from a surface, we can talk about the pressure of a fluid element (say a tiny cube) -- and pressure would act normal to the six sides of the cube.
The other kind of aerodynamic force is shear.  Shear acts tangent to a surface.  For the fluid element, it acts parallel (tangent) to the six faces of the cube.
The total aerodynamic force is the integral of the pressure and shear forces on a body.
Usually, the shear forces are insignificant in lift -- so the integral of pressures would be the dominant component of the lift force.
In particular, if you are doing an inviscid analysis, the shear forces will be exactly zero and the integral of the pressure is the forces.

This does not answer what the vectors in the picture represent. I'm 100% certain you have seen a picture like that before.

The vectors represent pressure.  This is likely an inviscid analysis -- or only a plot of pressure (ignoring shear).  If the pressures are multiplied by tiny differential areas on the surface, then they are tiny force vectors.  Pressure and force look the same.
Pressure is relative to a reference (freestream, vacuum, or whatever).  This is why we usually only talk about difference in pressure.  If you change the reference value, all the arrows will be elongated or shortened by the same amount. Applying the shift all around the airfoil cancels out, so the integrated force is the same.

The image doesn't convey the entire picture.
The main force is Newtonian action/reaction. A wing drives air down, more or less "planing" on it (hence, airplane), like a boat on the step. A secondary force is the wing being more or less sucked upward by the low pressure zone on top.
But mainly, the wing is inducing a large package of air, mostly above and some below, to move downward as it passes. The result is downwash, the final result of the Newtonian action/reaction.  You can make a wing from a flat plate, but the lift it can generate will be limited mostly to what air can be directly deflected mechanically by the plate's angle to the flow (the boat planing on water effect). Lots of light planes with steel tube structures have tails that are basically flat plates, and get by on air deflection alone for stabilizing and maneuvering forces, because the lift coefficients don't need to be very high.
On a normal airfoil, the curved surfaces and resulting pressure distribution that ends up with suction on top, in addition to the wing to wanting to move into the low pressure, allows a much larger package of air, well above the wing, to be induced to move downward than by direct mechanical deflection, and does this across a range of angles to the flow. The affected area above the wing, that is, the zone containing molecules that ended up lower than they were before the wing passed and ended up as part of the wing's downwash, may extend several chord lengths above.

We have a simple fact that can resolve the question in one go -
A gas (e. g. air) is physically unable to "pull" something.
Solids are capable of pulling (i.e. experience negative pressures). Liquids can do some pulling based on their surface tension. But a gas always exerts a positive pressure, it "pushes".
Whith this in mind, we can only conclude that the wing is pushed from below.

Very simple wind tunnel tests show that the pressure over a lifting airfoil is lower than the ambient one, see for example this one...

The integral sum of the pressure forces is what matters.


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Constant Speed Propeller and Angle of Attack
Question regarding the AoA (Angle of Attack) on a Constant speed.
The CSU (governor) sets the blade angle based on the desired RPM, and the flyweights keep it from speeding up or slowing down using the the spring preload that resists the flyweights.
So it sets the blade pitch.
What about the AoA?
E.g. on a run-up (no forward motion), the constant speed prop would speed up to a certain RPM, what happens to the AoA and why?
What about when accelerating down the runway? In this case the relative airflow would change angle, what about the AoA?
Many thanks!

FYI, some constant speed units are hydraulically driven.  Others are even electric.  I don't think you actually care about the mechanism that does the changing.
As airspeed changes -- whether in flight or running down the runway, the constant speed prop is adjusted to maintain constant RPM.  In fact, the main thing it responds to is changing airspeed.
You are perhaps forgetting the other primary engine control -- the throttle.  The throttle sets (roughly speaking) the power output of the engine.
The constant speed propeller adjust the prop such that it absorbs the engine's power at the desired speed.
Consider for a moment a fixed pitch prop.  For a fixed geometry prop (at a fixed airspeed and altitude), the power absorbed by the prop goes with the cube of RPM.
If you put an airplane with a fixed pitch prop into a dive, the airspeed will increase.  If you don't reduce the throttle, the RPM will increase -- perhaps to the point where you could over-speed the engine.  On an aircraft with a fixed pitch prop, the pilot manually adjusts the throttle to keep the engine/propeller within the safe operating range of the engine.
Back to a constant speed prop...  If you put an aircraft with a constant speed prop into a dive, the airspeed will increase.  The propeller un-loads from increased airspeed and would want to increase RPM.  However, the constant speed governor increases pitch to prevent RPM from increasing.  So, the constant speed unit helps the pilot safely operate the engine by preventing over-speeding.
Throughout all of these processes, the realized blade angle of attack can be whatever it needs to be.  It is certainly not constant.  It is also not at the point for maximum efficiency.  A constant speed propeller does not achieve maximum efficiency.  A CS prop has nothing to do with efficiency -- it has everything to do with power.


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Under part 103.7 how much more can an all composite amphibian weigh rather than 254 lbs since it already has a boat monocoque hull and sponsons?
Under part 103.7 how much more can an all composite amphibian weigh  rather than 254 lbs since it already has a boat monocoque hull and sponsons?